Question #03e18

1 Answer
Jul 18, 2017

#"8 g CaCl"_2#


For starters, you need to know exactly how many moles of chloride anions are present in your solution.

Since you know that the chloride anions have a concentration of #"0.500 mol L"^(-1)#, you can say that your sample will contain

#300 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.500 moles Cl"^(-)/(1color(red)(cancel(color(black)("L solution")))) = "0.15 moles Cl"^(-)#

Now, calcium chloride is soluble in water, which means that it dissociates completely in aqueous solution to produce calcium cations and chloride anions.

#"CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"Cl"_ ((aq))^(-)#

Notice that every mole of calcium chloride that dissociates produces #color(red)(2)# moles of chloride anions.

This means that in order for the solution to contain #0.15# moles of chloride anions, it must dissolve

#0.15 color(red)(cancel(color(black)("moles Cl"^(-)))) * "1 mole CaCl"_2/(color(red)(2)color(red)(cancel(color(black)("moles Cl"^(-))))) = "0.075 moles CaCl"_2#

Finally, to convert this to grams, use the molar mass of calcium chloride

#0.075 color(red)(cancel(color(black)("moles CaCl"_2))) * "110.98 g"/(color(red)(cancel(color(black)("mole CaCl"_2)))) = color(darkgreen)(ul(color(black)("8 g")))#

The answer must be rounded to one significant figure, the number of sig figs you have for the volume of the solution.